Question: Simplify and expand the following expression: $ \dfrac{4}{4a - 40}- \dfrac{1}{2a + 4}- \dfrac{2}{a^2 - 8a - 20} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{4}{4a - 40} = \dfrac{4}{4(a - 10)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{1}{2a + 4} = \dfrac{1}{2(a + 2)}$ We can factor the quadratic in the third term: $ \dfrac{2}{a^2 - 8a - 20} = \dfrac{2}{(a - 10)(a + 2)}$ Now we have: $ \dfrac{4}{4(a - 10)}- \dfrac{1}{2(a + 2)}- \dfrac{2}{(a - 10)(a + 2)} $ The least common multiple of the denominators is: $ 8(a - 10)(a + 2)$ In order to get the first term over $8(a - 10)(a + 2)$ , multiply by $\dfrac{2(a + 2)}{2(a + 2)}$ $ \dfrac{4}{4(a - 10)} \times \dfrac{2(a + 2)}{2(a + 2)} = \dfrac{8(a + 2)}{8(a - 10)(a + 2)} $ In order to get the second term over $8(a - 10)(a + 2)$ , multiply by $\dfrac{4(a - 10)}{4(a - 10)}$ $ \dfrac{1}{2(a + 2)} \times \dfrac{4(a - 10)}{4(a - 10)} = \dfrac{4(a - 10)}{8(a - 10)(a + 2)} $ In order to get the third term over $8(a - 10)(a + 2)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{2}{(a - 10)(a + 2)} \times \dfrac{8}{8} = \dfrac{16}{8(a - 10)(a + 2)} $ Now we have: $ \dfrac{8(a + 2)}{8(a - 10)(a + 2)} - \dfrac{4(a - 10)}{8(a - 10)(a + 2)} - \dfrac{16}{8(a - 10)(a + 2)} $ $ = \dfrac{ 8(a + 2) - 4(a - 10) - 16} {8(a - 10)(a + 2)} $ Expand: $ = \dfrac{8a + 16 - 4a + 40 - 16}{8a^2 - 64a - 160} $ $ = \dfrac{4a + 40}{8a^2 - 64a - 160}$ Simplify: $ = \dfrac{a + 10}{2a^2 - 16a - 40}$